Question

The area of the smaller region enclosed by the curves $y^2=8 x+4$ and $x^2+y^2+4 \sqrt{3} x-4=0$ is equal to

• A. $\frac{1}{3}(2-12 \sqrt{3}+8 \pi)$
• B. $\frac{1}{3}(2-12 \sqrt{3}+6 \pi)$
• C. $\frac{1}{3}(4-12 \sqrt{3}+8 \pi)$
• D. $\frac{1}{3}(4-12 \sqrt{3}+6 \pi)$

Answer 1

Answer: (C)

$x^2+y^2+4 \sqrt{3} x-4=0$
$ y^2=8 x+4$
Point of intersections are $(0,2) \&(0,-2)$ Both are symmetric about $x$-axis Area $=2 \int\limits_0^2\left(\sqrt{16-y^2}-2 \sqrt{3}\right)-\left(\frac{y^2-4}{8}\right) d y$ On solving Area $=\frac{1}{3}[8 \pi+4-12 \sqrt{3}]$